Phonon Statistics
• Crystal vibrations: There are two kinds of “quantization”;
The limitation of the allowed frequencies in a finite crystal to the normal modes by the boundary conditions. (classical atoms and classical waves;, only specific wavelengths, m2L, and their corresponding frequencies, m, are allowed)
The limitation of the allowed energies in one of specific mode of an oscillator. (quantum mechanical; we consider the individual atoms are now “particles” with wave-like properties. The energy in a vibration with a particular allowed mode (state), n, must be of the form
Both limitations must be satisfied by the vibration in a crystal, i.e.,
𝐸 𝑛 1
2 ℏ𝜔
𝐸 𝑛 ℏ𝜔𝑚
:The energy corresponding to the nth state of the mth mode.
Phonon statics
• What is the average number of phonons with frequency ω
mat temperature T?
(N possible modes – ω
mis a particular mode)
If 𝑥 1, 𝑥 1 1 𝑥
𝑥 𝑒 , 𝑥 𝑒 ℏ
𝑃 𝑒 1 𝑒 ℏ
𝐸 𝑛ℏ𝜔 : Energy corresponding to 𝑛th state of 𝑚th mode, i.e., the state with 𝑛 phonons, each with energy of ℏ𝜔
𝑃 𝑒 /
∑ 𝑒 /
(1) 𝑃𝑚𝑛 = probability of exciting the state with 𝐸𝑚𝑛 in the crystal at temperature T
Phonon statics
(2) average energy of the mth mode
(3) average number of phonon with ω
m𝐸 𝑃 𝐸
𝑛ℏ𝜔 𝑒 ℏ 1 𝑒 ℏ 𝐸 zero point energy
𝑛𝑥 𝑛𝑥 𝑥
1 𝑥
𝐸 𝐸 ℏ𝜔
𝑒ℏ / 1
𝑛̄ 𝐸 𝐸
ℏ𝜔
𝑛̄ 1
𝑒ℏ 1
Bose Einstein distribution
: Applicable to both phonons and photons
Phonon statics
1
𝑒 / 1
Bose-Einstein
Boltzman 𝑒
If ℏ𝜔 ≫ kT, the Bose- Einstein distribution reduces to the simple Boltzmann distribution.
E
If ℏ 1 (large T), 𝐸 𝐸 ℏ𝜔 1 ℏ𝜔
𝑘𝑇 1 𝑘𝑇
1
2𝑘𝑇 : kinetic energy, 1
2𝑘𝑇: potential energy
Total energy (there are 3N total number of modes in three dimensions)
𝐸 𝐸 3𝑁𝑘𝑇
The Hydrogen atom
• A single electron bound to a single proton by a Coulomb attractive force.
• An understanding of the solution for the hydrogen atom allow at least a qualitative understanding of elementary atomic structure.
sin cos sin sin cos
x r y r z r
Spherical coordinate system
The Hydrogen atom
• The Schrödinger equation for the hydrogen atom is
where the potential energy is Coulomb attraction between postive proton and negative electron.
• For spherical coordinates,
• Solutions by seeking a separation of the variables
1 𝑟
𝛿
𝛿𝑟 𝑟 𝛿
𝛿𝑟 𝜓 1
𝑟 sin 𝜃 𝛿
𝛿𝜃 sin 𝜃 𝛿
𝛿𝜃 𝜓 1
𝑟 sin 𝜃
𝜕 𝜓
𝜕𝜙
2𝑚 ℏ
𝑞
4𝜋𝜀 𝑟 𝐸 𝜓 0
∇ 𝜓 2𝑚
ℏ E V 𝑟 𝜓 0
V 𝑟 𝑞
4𝜋𝜀 𝑟
𝜓 𝑟, 𝜃, 𝜙 𝑅 𝑟 Θ 𝜃 Φ 𝜙
:central force field (only a function of distance from the center of force and independent of the particular direction)
V 𝑟
The Hydrogen atom
• The resulting differential equations for each of the variables are
• Angular parts [the equations for () & (θ)] are independent of V(r) and are the same for all central force-field systems.
: expressible in “spherical harmonics” and do not affect the allowed energies for the free hydrogen atom.
• Quantization is expected because an electron is confined physically. But, there are no geometric boundary conditions.
• However, the requirement is that the wave function should be mathematically well behaved.
• There are three quantum numbers because there are three kinds of variables.
1 Φ
𝜕 Φ
𝜕𝜙 𝐴 1
sin 𝜃 𝑑
𝑑𝜃 sin 𝜃𝑑Θ
𝑑𝜃 𝐵 𝐴
sin 𝜃 Θ 0 𝑑 𝑅
𝑑𝑟
2 𝑟
𝑑𝑅 𝑑𝑟
2𝑚
ℏ 𝐸 𝑉 𝑟 𝐵
𝑟 𝑅 0
(A & B are constants)
The solutions of Φ(ϕ), Θ(θ)
• Φ(ϕ)
In order for Φ(ϕ) to be single valued, Φ(ϕ+2π)= Φ(ϕ)
E in a free hydrogen atom does not depend on ml unless magnetic field is present.
(Zeeman effect)
• Θ(θ)
Θ(θ) involves polynomials in sin θ and cos θ.
For Θ(θ) to be finite, a polynomial must be terminated after a finite number of terms.
The termination condition introduces another quantum number ℓ such that
The total angular momentum for a given state of the hydrogen is
1 1
2 2
1 2
iA iA
c e c e
1, 2, :
( 1) 0,
l
B m
angular momentum quantum number
12 , 0, 1, 2 : the magnetic
l l
l
A m m
m quantum number
( 1)
12
The solutions of Φ(ϕ), Θ(θ)
• Θ(θ)
2 /2 2
1 (cos 1)
( ) (cos ) (1 cos )
2 ! (cos )
l l l
l
l m l
m m
l l l m
P d
l d
Spectroscopic
designation l ml 𝑃 cos𝜃
s 0 0 1 1
p 1 0 cos 1
p 1 1 sin exp(i
d 2 0 (3cos21)/2 1
d 2 1 3sincos exp(i
d 2 2 3sin2 exp(i2
[𝑎𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑒𝑑 𝐿𝑒𝑔𝑒𝑛𝑑𝑟𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠]
The solution for R(r)
• For R(r), the differential equation becomes
• The solution of the above equation can once again be written in the form of a polynomial that must be terminated after a finite number of terms in order for R(r) to be finite and well behaved.
• The termination conditions introduce the third quantum
number, n, so called the principal quantum number, because the energy depends only on the value of n in the hydrogen atom.
𝑑 𝑅 𝑑𝑟
2 𝑟
𝑑𝑅 𝑑𝑟
2𝑚
ℏ 𝐸 𝑞
4𝜋𝜀 𝑟
ℓ ℓ 1
𝑟 𝑅 0
The solution for R(r)
• The radial wave functions have the form
• Combine the various portions of the wave functions:
2
, 2
1
( ) ( 2 )
1( ) ( )
n n
n n
R r exp d e p x
d
d e
d
ℓ 𝑛 1 𝑛 1, 2, ⋯ 𝜌 2𝑟/𝑛𝑎
𝑎 4𝜋𝜀 ℏ /𝑚𝑞
,
, , ,
( )
,( ) ( )
m ms m
n
R
nr
m
: Bohr radius - the radius of the circular orbit of the lowest energy state (ground state)
𝐿𝑎𝑔𝑢𝑒𝑟𝑟𝑒 𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙𝑠
Energy of the hydrogen atom
• Allowed energies in the hydrogen atom
m is the reduced mass of the hydrogen atom, which is defined as . Since , the reduced mass of the hydrogen atom reduces to . The difference between the reduced mass and the electron mass is 0.05%
Ground state of the hydrogen atom corresponds to an energy of , called 1 Rydberg.
When an electron in energy state undergoes a transition to the ground state, the energy difference is emitted as a photon.
𝐸 𝑚𝑞
32𝜋 𝜀 ℏ ⋅ 1 𝑛
( )
N e N e
m m m m m
N m
em
e1 13.5 E eV
En
(En E1)
4
1 2 2
( ) 1 1
n n 2
E E E mq
n
“Lyman series” of optical emission lines for hydrogen atm
Energy of the hydrogen atom
Electron Spin
• In addition to the orbital angular momentum described by ℓ, the electron also has a kind of intrinsic angular momentum called “electron spin”.
• The treatment of the spin angular momentum is closely analogous to the treatment of orbital angular momentum.
• The z(arbitrary) component of the spin angular momentum
The spin angular momentum vector 𝑆
𝑆 𝑆 𝑠 𝑠 1 ℏ 𝑠: spin quantum number
𝑆 𝑚 ℏ 𝑚 : spin quantum number
𝑚 1
1 2
2: spin up 1
2: spin down
Orbital of hydrogen atom
• Allowed energies depend only on n
• Example
① 1s: n = 1, ℓ = 0, 𝑚
ℓ= 0
𝐸 𝑚𝑞
32𝜋𝜀 ℏ
1 𝑛
x
y z
Θ(θ) and Φ(ϕ) are constants. → Spherical symmetry state
Two spin quantum numbers (m
s)
→ two s states with the same energy
Orbital of hydrogen atom
② p state:
n = 2, ℓ = 1, 𝑚
ℓ= -1, 0, +1, 𝑚 =
p states have 3 orbital states.
Due to spin, p states have 6 possible states.
All the states have the same energy in hydrogen atom.
When the states with different spatial distributions (set of quantum numbers) have the same energy, it is said to as “degenerate”.
Wave functions for the degenerate states:
𝜓 , , 𝑅 , 𝑟 sin 𝜃 𝑒 𝜓 , , 𝑅 , 𝑟 cos 𝜃 𝜓 , , 𝑅 , 𝑟 sin 𝜃 𝑒
Angular distribution of the p state, 𝜓 , , , is symmetry about the z axis. (Lobes of probability density)
Distribution of wave functions
(Probability density)
Pauli Exclusion Principle
• “No two electrons in the same system can have all the quantum numbers the same, n, ℓ, 𝑚 , 𝑚 .”
• “The wave function for any system of electrons must be anti- symmetric with respect to the interchange of any two
electrons.
: 2 for : 3 for : 5 for : 7 for
0 ( ) 1 (p) 2 (d) 3 (f) has two states, : 6 states,
ms
m m
s m
s p
: 10 states, : 14 states
d f
Pauli Exclusion Principle
1 2
2 1
2 2
H : 1 He : 1 Li : 1 2 Be : 1 2
s s s s s s
2 2
1 2 3 4 5
6
1 2 B : 2 C : 2 N : 2 O : 2 F : 2 Ne : 2
s s p p p p p
p
(2)
(2) 2 X 3 = 6
